Integral 1 to 2 of (x^3)lnxdx

What do we use as u? I tried lnx and x^3 and it didnt seem to work.

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- Mar 4th 2013, 01:53 PMSteelers72Definite integral with u substitution
Integral 1 to 2 of (x^3)lnxdx

What do we use as u? I tried lnx and x^3 and it didnt seem to work. - Mar 4th 2013, 02:39 PMBobPRe: Definite integral with u substitution
The method is to integrate by parts.

- Mar 4th 2013, 05:26 PMhollywoodRe: Definite integral with u substitution
You should remember the acronym LIATE: you want to choose for u whatever comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. In this case you would choose $\displaystyle u=\ln{x}$.

- Hollywood - Mar 4th 2013, 08:49 PMSteelers72Re: Definite integral with u substitution
- Mar 4th 2013, 11:36 PMsbhatnagarRe: Definite integral with u substitution
Here's another method:

$\displaystyle \int_1^2 x^a dx=\frac{2^{1+a}-1}{1+a}$

Differentiate with respect to $\displaystyle a$:

$\displaystyle \int_1^2 x^a \log(x) dx=\frac{2^{1+a}\log(2)(1+a)-2^{1+a}+1}{(1+a)^2}$

Putting $\displaystyle a=3$, gives the answer

$\displaystyle \int_1^2 x^3 \log(x) dx= \frac{4\times 2^4 \log(2)-2^4+1}{4^2}=4\log(2)-\frac{15}{16}$